Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
RonPurewal
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Re: Combination Manhattan CAT #5 problem 31

by RonPurewal Wed Nov 12, 2014 4:30 am

more importantly—

if you're messing around with expressions that look like "3! – 2(2!)", here are some thoughts that should occur to you:
• 3! is only 6.
• so, this is even less than 6.
• if there are that few possibilities, then, the heck with these crazy calculations—let's just count.

if you have the 2 daughters in the back seat (plus one last person), then it's a VERY short list: they have to occupy the two window seats, so there are only two possibilities.
so, for the first term above, you can just write 2•2•2. no need to mess around with weird factorial expressions.
RAHULZ400
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Re: Combination Manhattan CAT #5 problem 31

by RAHULZ400 Fri Sep 30, 2016 10:41 pm

I have attempted this question in the following manner:

Case1: 2 ways for parents to sit on front seat
2 ways for a daughter to sit on next seat
3 ways for (daughter) for any back seat
2 ways for son for left seats
1 way for parent (left over)

total ways = 2*2*3*2 = 24

Case 2: 2 ways for parent for 1st seat
2 ways for other parent or son to grab the next front seat
2 ways for daughter to sit on 1st back seat
1 way for daughter to sit on 3rd back seat
1 way left for either the left over parent or son

total ways = 2*2*2 = 8 ways

Hence total number of combinations possible = 24+8 = 32

Is this approach fine. Please let me know your feedback.
RonPurewal
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Re: Combination Manhattan CAT #5 problem 31

by RonPurewal Fri Oct 07, 2016 2:31 pm

Case1: 2 ways for parents to sit on front seat
2 ways for a daughter to sit on next seat
3 ways for (daughter) for any back seat
2 ways for son for left seats
1 way for parent (left over)

total ways = 2*2*3*2 = 24


^^ this works.
you CAN simplify it a bit, as follows:
• pick a parent to drive (2 possibilities)
• pick a daughter for the other front seat (2 possibilities)
• the last 3 people can be arranged freely (3! = 3·2·1 possibilities)
thus 2·2·3·2·1

this is basically the same thing you already have, except you don't have to think explicitly through the 3·2·1 part as you did—you can just do that as a factorial.

Case 2: 2 ways for parent for 1st seat
2 ways for other parent or son to grab the next front seat
2 ways for daughter to sit on 1st back seat
1 way for daughter to sit on 3rd back seat
1 way left for either the left over parent or son

total ways = 2*2*2 = 8 ways


^^ yep.